a 7 ae 
? 
7 aie < 
ay . 
ae = 
‘ : BEAMS AND BENDING 97 
te for M, is a parabola whose axis LN bisects AH at right angles. 
L ae . Putting ants in the equation M,= =e (l-x-a), 
“- N. eb _We- ~ay 
MINS 4l 
Tn like manner it can be shown that the curve for M, is a parabola 
KOB whose axis PO bisects KB at right angles, AK being equal to b, 
an¢ . These two parabolas intersect at a point S on the 
rtical through E, which divides the field of W, from the field of W,, 
1 the beading moment diagram for the whole girder i is ANSB. 
“A parabola, whose axis is the vertical through C, the middle of the 
van, and which touches the larger of the two parabolas at the base line, 
vi be the bending moment curve for the equivalent uniform dead load. 
et'ANH be the larger of the two parabolas ANH and KOB. Produce 
» axis LN to T, making NT'=LN, then AT is the tangent to the 
4. ANH at A. AT will also be the tangent to the circumscribing 
abola at A. Produce AT to meet the vertical CQ at U. Bisect QU 
ity, then V is the vertex of the circumscribing parabola. 
Tt_is easy to show that QV: LN:: AQ: AL, and therefore that the 
point V may be found by joining AN and producing to meet CQ at V. 
LN xAQ W(/-a)?_ 1 l-a_W(i-a) 
3 a ae OAS eye 
Tf the equivalent uniform dead load is w pet unit of length, then 
wl? - a) W(l =a) 
= , therefore ws as ee 
E When both loads are to the right of D the positive shearing force at 
T ales and as the loads move towards D, R, increases, and is greater the 
is toD. When W, passes to the left of D, the positive shearing 
aD is suddenly diminished by the amount W,, and is then equal 
to to F ee, but as W, moves to the left, W, being still to the right of D, 
R, increases, and therefore R, — W, i increases until W, is at D. When 
be oth loads are to the left of DD there is no positive shearing force at D. 
Hence the positive shearing force at D is a maximum, either when W, is 
at it D the righ hee W, is at D, or, more correctly, when W, or Wy is just. to 
to 
- Placing the loads so that W, is at D, or just to the right -* D, as shown 
a (0) Fig. 125, the positive shearing force at D =F, =R, = > AS aa), 
Placing the loads so that W, is at D, or just to the right of D, as shown 
at), the positive shearing force at D =F, = R, - W, = =7¢ -x+b)-W,. 
3 F, will be greater than F, when W(l-2+))-W,> vO -“%-a), 
;. that is, when W >/, and F, will be less than F, when W. <i. 
f G 
